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Math Help - Trigonometric Substitution

  1. #1
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    Trigonometric Substitution

    integrate: \frac{x^2}{\sqrt{4-9x^2}}


    i worked it down to this, first off is this right so far. And how would i simplify this so that its integrable. also how do you do the integrate sign in latex? thx guys.

    integrate: \frac{x^2}{\sqrt{4-9x^2}} = integrate: \frac{\frac{2}{3}(cost)^2}{2sint} * (-\frac{2}{3} sint) dt
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  2. #2
    Super Member Deadstar's Avatar
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    Quote Originally Posted by x5pyd3rx View Post
    integrate: \frac{x^2}{\sqrt{4-9x^2}}


    i worked it down to this, first off is this right so far. And how would i simplify this so that its integrable. also how do you do the integrate sign in latex? thx guys.

    integrate: \frac{x^2}{\sqrt{4-9x^2}} = integrate: \frac{\frac{2}{3}(cost)^2}{2sint} * (-\frac{2}{3} sint) dt
    I'm guessing you set x = \frac{2}{3}\cos(t) yes?

    That should be fine (although personally I would have chosen sin since it's derivative is positive...

    But anyway...

    x = \frac{2}{3}\cos(t)

    dx = -\frac{2}{3}\sin(t)dt

    Integral becomes...

    \int -\frac{2}{3}\frac{\tfrac{4}{9}\cos^2(t)}{\sqrt{4-4\cos^2(t)}}\sin(t) dt.

    = \int -\frac{8\cos^2(t)}{54\sqrt{1-\cos^2(t)}}\sin(t)  dt

    = \int -\frac{4\cos^2(t)}{27\sin(t)}\sin(t)  dt

    = -\frac{4}{27} \int \cos^2(t)  dt

    Solve from here, sub back in what t equals after you've integrated.
    Last edited by Deadstar; April 20th 2010 at 11:21 AM.
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  3. #3
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    Quote Originally Posted by x5pyd3rx View Post
    integrate: \frac{x^2}{\sqrt{4-9x^2}}


    i worked it down to this, first off is this right so far. And how would i simplify this so that its integrable. also how do you do the integrate sign in latex? thx guys.

    integrate: \frac{x^2}{\sqrt{4-9x^2}} = integrate: \frac{\frac{2}{3}(cost)^2}{2sint} * (-\frac{2}{3} sint) dt


    \frac{x^2}{\sqrt{4-9x^2}}=\frac{x^2}{2\sqrt{1-\left(\frac{3x}{2}\right)^2}} , so now substittute \frac{3}{2}\,x=\sin u\Longrightarrow \frac{3}{2}\,dx=\cos u\,du , and:

    \int \frac{x^2}{\sqrt{4-9x^2}}\,dx=\int\frac{x^2}{2\sqrt{1-\left(\frac{3x}{2}\right)^2}}\,dx =\frac{1}{2}\int\frac{4/9\,\,\sin^2u\,\,(2/3)\,\cos udu}{\cos u}=\frac{4}{27}\int \sin^2u\,du =\frac{2}{27}\,(u-\sin u\cos u)+C

    If you want now put back everything in terms of u instead of x.

    Tonio
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  4. #4
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    Thx so much guys you both where a bunch of help
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